(1) The function $f(x)=\sin ^{2}x+2 \sqrt {3}\sin (x+ \frac {π}{4})\cos (x- \frac {π}{4})-\cos ^{2}x- \sqrt {3}$ can be simplified as follows:
$f(x)=-\cos 2x+2 \sqrt {3}\cdot( \frac { \sqrt {2}}{2}\sin x+\frac { \sqrt {2}}{2}\cos x)\cdot( \frac { \sqrt {2}}{2}\cos x+\frac { \sqrt {2}}{2}\sin x)- \sqrt {3}$
$f(x)=-\cos 2x+2 \sqrt {3}\cdot( \frac {1}{2}+\frac {1}{2}\sin 2x)- \sqrt {3}$
$f(x)= \sqrt {3}\sin 2x-\cos 2x=2\sin (2x- \frac {π}{6})$
Let $2kπ+ \frac {π}{2}\leqslant 2x- \frac {π}{6}\leqslant 2kπ+ \frac {3π}{2}$, we get $kπ+ \frac {π}{3}\leqslant x\leqslant kπ +\frac {5π}{6}$. Thus, the intervals where the function is decreasing are $[kπ+ \frac {π}{3},kπ +\frac {5π}{6}]$, where $k\in\mathbb{Z}$.

(2) On the interval $[- \frac {π}{12}, \frac {25}{36}π]$, we have $2x- \frac {π}{6}\in[- \frac {π}{3}, \frac {11π}{9}]$. The maximum value of $f(x)$ occurs when $2x- \frac {π}{6}= \frac {π}{2}$. Therefore, the maximum value of the function $f(x)$ is $\boxed{2}$.